Solution

A spherical planet of radius \(a\) has a variable density \(f(r)\) which depends only on the distance \(r\) from the planet’s centre. Show that the average density of the planet is \[3 \int_0^1 t^2 f(at) \:dt.\]

Alternatively, \(\delta V =\) surface area of the hollow sphere\(\times\) its thickness.

We know that the density of this hollow sphere is \(f(r)\), so the mass of the hollow sphere is \(4\pi r^2 f(r) \delta r\).

Taking the limit, the total mass of the whole sphere is therefore

\[\int_0^a 4\pi r^2 f(r) \:dr.\]

We know that the volume of the whole sphere is \(\dfrac{4\pi}{3}a^3\), so the average density of the whole sphere is \[\dfrac{\int_0^a 4\pi r^2 f(r) \:dr}{\frac{4\pi}{3}a^3}\] which is \[\int_0^a \dfrac{3 r^2 f(r)}{a^3} \:dr.\]

If we now put \(r = at\), we have \(dr = a\: dt\), and substituting in, we find that the average density is \[3 \int_0^1 t^2 f(at) \:dt,\] as required.

Find the average density correct to two significant figures in each of the three cases:

Alternatively, rather than using partial fractions, we could write \(u=1+t\) so that our integral becomes \[3\int_1^2 \frac{(u-1)^3}{u^3}\:du\] which we can expand, cancel and integrate as powers of \(u\) with the same result as above.